Chapter 12.2, Problem 1OE

To determine

Find the values of the six trigonometric functions of $\theta$ .

Answer to Problem 1OE

  $\sin \theta =\frac{7}{\sqrt{58}}$

  $\cos \theta =\frac{3}{\sqrt{58}}$

  $\tan \theta =\frac{7}{3}$

  $\csc \theta =\frac{\sqrt{58}}{7}$

  $\sec \theta =\frac{\sqrt{58}}{3}$

  $\cot \theta =\frac{3}{7}$

Explanation of Solution

Given:

  

Formula Used:

  $\sin \theta =\frac{\text{length of side opposite to }\theta }{\text{length of the hypotenuse}}$

  $\cos \theta =\frac{\text{length of side adjacent to }\theta }{\text{length of the hypotenuse}}$

  $\tan \theta =\frac{\text{length of side opposite to }\theta }{\text{length of side adjacent to }\theta }$

  $\csc \theta =\frac{1}{\sin \theta }$

  $\sec \theta =\frac{1}{\cos \theta }$

  $\cot \theta =\frac{1}{\tan \theta }$

Calculation:

In the triangle ,

x = length of side adjacent to $\theta$ = 3 units

y = length of side opposite to $\theta$= 7 units

r = length of the hypotenuse = $\sqrt{x^2+y^2}=\sqrt{9+49}=\sqrt{58}$

So,

  $\sin \theta =\frac{\text{length of side opposite to }\theta }{\text{length of the hypotenuse}}=\frac{y}{r}=\frac{7}{\sqrt{58}}$

  $\cos \theta =\frac{\text{length of side adjacent to }\theta }{\text{length of the hypotenuse}}=\frac{x}{r}=\frac{3}{\sqrt{58}}$

  $\tan \theta =\frac{\text{length of side opposite to }\theta }{\text{length of side adjacent to }\theta }=\frac{y}{x}=\frac{7}{3}$

  $\csc \theta =\frac{1}{\sin \theta }=\frac{\sqrt{58}}{7}$

  $\sec \theta =\frac{1}{\cos \theta }=\frac{\sqrt{58}}{3}$

  $\cot \theta =\frac{1}{\tan \theta }=\frac{3}{7}$